$h(x)=x^8$ $h'(x)=$
Explanation: $h$ is of the form $x^n$ and therefore we can apply the power rule: $\dfrac{d}{dx}[x^n]=n\cdot x^{n-1}$ $\begin{aligned} h'(x)&=\dfrac{d}{dx}[x^{{8}}] \\\\ &={8}x^{{8}-1} \\\\ &=8x^7 \end{aligned}$ In conclusion, $h'(x)=8x^7$